# 5.4 Comparison tests  (Page 3/7)

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## Limit comparison test

Let ${a}_{n},\phantom{\rule{0.2em}{0ex}}{b}_{n}\ge 0$ for all $n\ge 1.$

1. If $\underset{n\to \infty }{\text{lim}}{a}_{n}\text{/}{b}_{n}=L\ne 0,$ then $\sum _{n=1}^{\infty }{a}_{n}$ and $\sum _{n=1}^{\infty }{b}_{n}$ both converge or both diverge.
2. If $\underset{n\to \infty }{\text{lim}}{a}_{n}\text{/}{b}_{n}=0$ and $\sum _{n=1}^{\infty }{b}_{n}$ converges, then $\sum _{n=1}^{\infty }{a}_{n}$ converges.
3. If $\underset{n\to \infty }{\text{lim}}{a}_{n}\text{/}{b}_{n}=\infty$ and $\sum _{n=1}^{\infty }{b}_{n}$ diverges, then $\sum _{n=1}^{\infty }{a}_{n}$ diverges.

Note that if ${a}_{n}\text{/}{b}_{n}\to 0$ and $\sum _{n=1}^{\infty }{b}_{n}$ diverges, the limit comparison test gives no information. Similarly, if ${a}_{n}\text{/}{b}_{n}\to \infty$ and $\sum _{n=1}^{\infty }{b}_{n}$ converges, the test also provides no information. For example, consider the two series $\sum _{n=1}^{\infty }1\text{/}\sqrt{n}$ and $\sum _{n=1}^{\infty }1\text{/}{n}^{2}.$ These series are both p -series with $p=1\text{/}2$ and $p=2,$ respectively. Since $p=1\text{/}2>1,$ the series $\sum _{n=1}^{\infty }1\text{/}\sqrt{n}$ diverges. On the other hand, since $p=2<1,$ the series $\sum _{n=1}^{\infty }1\text{/}{n}^{2}$ converges. However, suppose we attempted to apply the limit comparison test, using the convergent $p-\text{series}$ $\sum _{n=1}^{\infty }1\text{/}{n}^{3}$ as our comparison series. First, we see that

$\frac{1\text{/}\sqrt{n}}{1\text{/}{n}^{3}}=\frac{{n}^{3}}{\sqrt{n}}={n}^{5\text{/}2}\to \infty \phantom{\rule{0.2em}{0ex}}\text{as}\phantom{\rule{0.2em}{0ex}}n\to \infty .$

Similarly, we see that

$\frac{1\text{/}{n}^{2}}{1\text{/}{n}^{3}}=n\to \infty \phantom{\rule{0.2em}{0ex}}\text{as}\phantom{\rule{0.2em}{0ex}}n\to \infty .$

Therefore, if ${a}_{n}\text{/}{b}_{n}\to \infty$ when $\sum _{n=1}^{\infty }{b}_{n}$ converges, we do not gain any information on the convergence or divergence of $\sum _{n=1}^{\infty }{a}_{n}.$

## Using the limit comparison test

For each of the following series, use the limit comparison test to determine whether the series converges or diverges. If the test does not apply, say so.

1. $\sum _{n=1}^{\infty }\frac{1}{\sqrt{n}+1}$
2. $\sum _{n=1}^{\infty }\frac{{2}^{n}+1}{{3}^{n}}$
3. $\sum _{n=1}^{\infty }\frac{\text{ln}\left(n\right)}{{n}^{2}}$
1. Compare this series to $\sum _{n=1}^{\infty }\frac{1}{\sqrt{n}}.$ Calculate
$\underset{n\to \infty }{\text{lim}}\phantom{\rule{0.2em}{0ex}}\frac{1\text{/}\left(\sqrt{n}+1\right)}{1\text{/}\sqrt{n}}=\underset{n\to \infty }{\text{lim}}\begin{array}{c}\frac{\sqrt{n}}{\sqrt{n}+1}\\ \phantom{\rule{0.2em}{0ex}}\end{array}=\underset{n\to \infty }{\text{lim}}\phantom{\rule{0.2em}{0ex}}\frac{1\phantom{\rule{0.2em}{0ex}}\text{/}\sqrt{n}}{1+1\text{/}\sqrt{n}}=1.$
By the limit comparison test, since $\sum _{n=1}^{\infty }\frac{1}{\sqrt{n}}$ diverges, then $\sum _{n=1}^{\infty }\frac{1}{\sqrt{n}+1}$ diverges.
2. Compare this series to $\sum _{n=1}^{\infty }{\left(\frac{2}{3}\right)}^{n}.$ We see that
$\underset{n\to \infty }{\text{lim}}\frac{\left({2}^{n}+1\right)\text{/}{3}^{n}}{{2}^{n}\text{/}{3}^{n}}=\underset{n\to \infty }{\text{lim}}\frac{{2}^{n}+1}{{3}^{n}}·\frac{{3}^{n}}{{2}^{n}}=\underset{n\to \infty }{\text{lim}}\frac{{2}^{n}+1}{{2}^{n}}=\underset{n\to \infty }{\text{lim}}\left[1+{\left(\frac{1}{2}\right)}^{n}\right]=1.$

Therefore,
$\underset{n\to \infty }{\text{lim}}\frac{\left({2}^{n}+1\right)\text{/}{3}^{n}}{{2}^{n}\text{/}{3}^{n}}=1.$

Since $\sum _{n=1}^{\infty }{\left(\frac{2}{3}\right)}^{n}$ converges, we conclude that $\sum _{n=1}^{\infty }\frac{{2}^{n}+1}{{3}^{n}}$ converges.
3. Since $\text{ln}\phantom{\rule{0.1em}{0ex}}n compare with $\sum _{n=1}^{\infty }\frac{1}{n}.$ We see that
$\underset{n\to \infty }{\text{lim}}\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}n\text{/}{n}^{2}}{1\text{/}n}=\underset{n\to \infty }{\text{lim}}\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}n}{{n}^{2}}·\frac{n}{1}=\underset{n\to \infty }{\text{lim}}\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}n}{n}.$

In order to evaluate $\underset{n\to \infty }{\text{lim}}\text{ln}\phantom{\rule{0.1em}{0ex}}n\text{/}n,$ evaluate the limit as $x\to \infty$ of the real-valued function $\text{ln}\left(x\right)\text{/}x.$ These two limits are equal, and making this change allows us to use L’Hôpital’s rule. We obtain
$\underset{x\to \infty }{\text{lim}}\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}x}{x}=\underset{x\to \infty }{\text{lim}}\frac{1}{x}=0.$

Therefore, $\underset{n\to \infty }{\text{lim}}\text{ln}\phantom{\rule{0.1em}{0ex}}n\text{/}n=0,$ and, consequently,
$\underset{n\to \infty }{\text{lim}}\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}n\text{/}{n}^{2}}{1\text{/}n}=0.$

Since the limit is $0$ but $\sum _{n=1}^{\infty }\frac{1}{n}$ diverges, the limit comparison test does not provide any information.
Compare with $\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}$ instead. In this case,
$\underset{n\to \infty }{\text{lim}}\frac{\text{ln}\phantom{\rule{0.2em}{0ex}}n\text{/}{n}^{2}}{1\text{/}{n}^{2}}=\underset{n\to \infty }{\text{lim}}\frac{\text{ln}\phantom{\rule{0.2em}{0ex}}n}{{n}^{2}}·\frac{{n}^{2}}{1}=\underset{n\to \infty }{\text{lim}}\text{ln}\phantom{\rule{0.1em}{0ex}}n=\infty .$

Since the limit is $\infty$ but $\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}$ converges, the test still does not provide any information.
So now we try a series between the two we already tried. Choosing the series $\sum _{n=1}^{\infty }\frac{1}{{n}^{3\text{/}2}},$ we see that
$\underset{n\to \infty }{\text{lim}}\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}n\text{/}{n}^{2}}{1\text{/}{n}^{3\text{/}2}}=\underset{n\to \infty }{\text{lim}}\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}n}{{n}^{2}}·\frac{{n}^{3\text{/}2}}{1}=\underset{n\to \infty }{\text{lim}}\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}n}{\sqrt{n}}.$

As above, in order to evaluate $\underset{n\to \infty }{\text{lim}}\text{ln}\phantom{\rule{0.1em}{0ex}}n\text{/}\sqrt{n},$ evaluate the limit as $x\to \infty$ of the real-valued function $\text{ln}\phantom{\rule{0.1em}{0ex}}x\text{/}\sqrt{x}.$ Using L’Hôpital’s rule,
$\underset{x\to \infty }{\text{lim}}\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}x}{\sqrt{x}}=\underset{x\to \infty }{\text{lim}}\frac{2\sqrt{x}}{x}=\underset{x\to \infty }{\text{lim}}\frac{2}{\sqrt{x}}=0.$

Since the limit is $0$ and $\sum _{n=1}^{\infty }\frac{1}{{n}^{3\text{/}2}}$ converges, we can conclude that $\sum _{n=1}^{\infty }\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}n}{{n}^{2}}$ converges.

Use the limit comparison test to determine whether the series $\sum _{n=1}^{\infty }\frac{{5}^{n}}{{3}^{n}+2}$ converges or diverges.

The series diverges.

## Key concepts

• The comparison tests are used to determine convergence or divergence of series with positive terms.
• When using the comparison tests, a series $\sum _{n=1}^{\infty }{a}_{n}$ is often compared to a geometric or p -series.

Use the comparison test to determine whether the following series converge.

$\sum _{n=1}^{\infty }{a}_{n}$ where ${a}_{n}=\frac{2}{n\left(n+1\right)}$

$\sum _{n=1}^{\infty }{a}_{n}$ where ${a}_{n}=\frac{1}{n\left(n+1\text{/}2\right)}$

Converges by comparison with $1\text{/}{n}^{2}.$

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